If $x \oslash y = x^{2}+4y^{2}$ and $x \triangleright y = x+3y$, find $(1 \oslash 3) \triangleright -1$.
Solution: First, find $1 \oslash 3$ $ 1 \oslash 3 = 1^{2}+4(3^{2})$ $ \hphantom{1 \oslash 3} = 37$ Now, find $37 \triangleright -1$ $ 37 \triangleright -1 = 37+(3)(-1)$ $ \hphantom{37 \triangleright -1} = 34$.